Question

1. a silver block, initially at 58.5°c, is submerged into 100.0 g of water at 24.8°c, in an insulated container. the final temperature of the mixture upon reaching thermal equilibrium is 26.2°c. what is the mass of the silver block?

Explanation:

Step1: Identify heat - transfer formula

The heat lost by the silver block is equal to the heat gained by the water. The heat - transfer formula is $q = mc\Delta T$, where $q$ is heat, $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the change in temperature. The specific heat capacity of water $c_{w}=4.184\ J/g\cdot^{\circ}C$ and the specific heat capacity of silver $c_{Ag}=0.235\ J/g\cdot^{\circ}C$.

Step2: Set up heat - balance equation

$q_{Ag}=-q_{w}$. Using the heat - transfer formula, we have $m_{Ag}c_{Ag}(T_{f}-T_{i,Ag})=-m_{w}c_{w}(T_{f}-T_{i,w})$.

Step3: Substitute known values

We know that $m_{w} = 100.0\ g$, $c_{w}=4.184\ J/g\cdot^{\circ}C$, $T_{i,w}=24.8^{\circ}C$, $T_{f}=26.2^{\circ}C$, $c_{Ag}=0.235\ J/g\cdot^{\circ}C$, and $T_{i,Ag}=58.5^{\circ}C$.
Substitute these values into the equation:
\[

$$\begin{align*} m_{Ag}\times0.235\times(26.2 - 58.5)&=- 100.0\times4.184\times(26.2 - 24.8)\\ m_{Ag}\times0.235\times(-32.3)&=-100.0\times4.184\times1.4\\ m_{Ag}\times(-7.5805)&=-585.76 \end{align*}$$

\]

Step4: Solve for $m_{Ag}$

\[m_{Ag}=\frac{- 585.76}{-7.5805}\approx77.3\ g\]

Answer:

$77.3\ g$